Exercises 31
Here, it turns out that there is only one possible choice for each of T, 1, and -':
T=1, 1_=0, --0=1, and-1 =0
Indeed, there is always only one possible choice (see Exercise 24 in Section 1.4). Thus, in
any boolean algebra, we may refer to T, 1, and each -x without ambiguity.
Solution. The proof requires showing that no matter what value x, y e B have, the ax-
ioms of a boolean algebra hold. As an example, we will show that the operation meet is
commutative. Let x, y E B. Then,
y x
V 0 1 v^0 1
X 0 0 1y 0 0 1
We can simply check that for all possible pairs, x V y = y V x for the meet operation.
Similar proofs are needed for the other axioms and will be left for the reader. N
In Section 2.1.4, we will consider this boolean algebra by another name. In place of 1
and 0, we will call the values of the elements TRUE and FALSE. The operations will be or
and and. Then, for example, we can interpret a variable x to be TRUE if there is a current
flowing in a wire X-and similarly, y to be TRUE if there is a current flowing in wire Y.
This turns out to be a very natural way to look at computer circuits.
- Let A ={1, 2, 3 ... , 10), B = {2, 3, 6, 81, and C = {3, 5, 4, 8, 2}. Find the follow-
ing:
(a) BUC
(b) BnC
(c) B - C
(d) A - B
(e) A - C
2. Let U={0,1,2,3,4,5, 6,7,8,9), A={0,1,2,3}, B={0,2,4}, and C=
{0, 3, 6, 9).
(a) FindAUB, AnB, A, (A n B), and (B U C) - A.
(b) Find P(A), P(B), 7P(A n B), P(A) n P(B).
(c) Is P(A U B) = P(A) U P(B)? Prove your answer.
(d) Why doesn't P(A) make sense?
3. Let A = {0, 3) and B = {x, y, z}. Find the following:
(a) A x B
(b) A x A x B
(c) B x A
(d) B x A x B
